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Thanks \Sigma\eta = F_{B} - F_{g} F_{B}(cos 75)(.08 m) = (2 kg)(9.8 m/s^2)(.33 m) F_{B} = \frac{(2 kg)(9.8 m/s^2)(.33 m)}{(cos 75)(.08 m)} = 312 N

A cook holds a 2 kg cartoon of milk at arm's length. What Force F_{B} must be excerted by the biceps muscle? (Ignore the weight of the forearm) See figure attached. \theta = 75 degrees Please help explain what I'm doing wrong. Correct answer is 312 N. \Sigma\eta = F(sin 75)(.08 m) - (2...

Thank you. My prof. told us to solve the question this way. He did mention that the equation he gave us for centripetal force was incorrect but he didn't tell the class why.

A roller-caster vechicle has a mass of 500 kg when fully loaded with passengers (Fig P7.28) (a) If the vechile has a speed of 20.0 m/s at point A, what is the force of the track on the vehicle at this point? (b) What is the maximum speed the vehicle can have at point B in order for gravity to...

A roller-caster vechicle has a mass of 500 kg when fully loaded with passengers (Fig p7.28) (a) If the vechile has a speed of 20.0 m/s at point A, what is the force of the track on the vehicle at this point? (b) What is the maximum speed the vehicle can have at point B in order for gravity to...

Thank you for checking my problem.

t = \frac{x}{v} = \frac{5.0 m}{2.46 m/s} = 2.03 s Corrected: t = \frac{x}{v} = \frac{5.0 m}{.081 m/s} = 62 s

I'm not sure if i did this problem correctly. Could someone check it over. A 730-N man stands in the middle of a frozen pond of a radius of 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2...

For the second projectile problem. You need to find the x and y vector components in order to solve for the amount of time. After you calculate the time you will be able to solve for the horizontal distance. v_ox = v_o cos\theta v_oy = v_o sin\theta \Delta y = v_oy t - \frac{1}{2}gt^2...

Need More Help Please. This is what i have so far. Part a: s = vot + (.5)at^2 s = 100m/s(3s) + .5(30m/s/s)(3s)^2 = 435 m y1 = s(sin 53) = 347 m y2= [100(sin53)^2 / 2(9.8)] = 325 m ytotal = 347 + 325 = 672 m correct answer is 1520 m The rest of the problem will be solved...

A rocket is launched at an angle of 53o above the horizontal with an initial speed of 100m/s. It moves for 3s along its initial line of motion with an accelration of 30m/s2. At this time its engines fail and the rocket proceeds to move as a free body. Find a)the maximum altitude reached by the...

Book The Aldrich library of FT-IR spectra